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779

answers:

5

Does C++ provide a guarantee for the lifetime of a temporary variable that is created within a function call but not used as a parameter? Here's an example class:

class StringBuffer
{
public:
    StringBuffer(std::string & str) : m_str(str)
    {
        m_buffer.push_back(0);
    }
    ~StringBuffer()
    {
        m_str = &m_buffer[0];
    }
    char * Size(int maxlength)
    {
        m_buffer.resize(maxlength + 1, 0);
        return &m_buffer[0];
    }
private:
    std::string & m_str;
    std::vector<char> m_buffer;
};

And here's how you would use it:

// this is from a crusty old API that can't be changed
void GetString(char * str, int maxlength);

std::string mystring;
GetString(StringBuffer(mystring).Size(MAXLEN), MAXLEN);

When will the destructor for the temporary StringBuffer object get called? Is it:

  • Before the call to GetString?
  • After GetString returns?
  • Compiler dependent?

I know that C++ guarantees that a local temporary variable will be valid as long as there's a reference to it - does this apply to parent objects when there's a reference to a member variable?

Thanks.

+15  A: 

The destructor for that sort of temporaries is called at the end of the full-expression. That's the most outer expression which is not part of any other expression. That is in your case after the function returns and the value is evaluated. So, it will work all nice.

It's in fact what makes expression templates work: They can keep hold references to that sort of temporaries in an expression like

e = a + b * c / d

Because every temporary will last until the expression

x = y

Is evaluated completely. It's quite concisely described in 12.2 Temporary objects in the Standard.

Johannes Schaub - litb
I never quite got around to getting a copy of the standard. I should make it a priority.
Mark Ransom
u can read about it here: http://www.kuzbass.ru:8086/docs/isocpp/special.html#class.temporary
Johannes Schaub - litb
+3  A: 

After the call to GetString returns.

David Segonds
A: 

StringBuffer is in the scope of GetString. It should get destroyed at the end of GetString's scope (ie when it returns). Also, I don't believe that C++ will guarantees that a variable will exist as long as there is reference.

The following ought to compile:

Object* obj = new Object;
Object& ref = &(*obj);
delete obj;
BigSandwich
I think I overstated the guarantee - it's for local temporaries only. But it does exist.
Mark Ransom
I've edited the question. Based on the answers supplied thus far, it seems to be a moot point however.
Mark Ransom
I still don't think your edit is correct: Object Object } //bad - will leave dangling reference.
BigSandwich
Mark Ransom
That's circular reasoning, but I feel like this discussion isn't really that important, so I'm not going to argue it.
BigSandwich
Mark Ransom
Thanks, for the link, that makes sense now.
BigSandwich
+1  A: 

litb's answer is accurate. The lifetime of the temporary object (also known as an rvalue) is tied to the expression and the destructor for the temporary object is called at the end of the full expression and when the destructor on StringBuffer is called, the destructor on m_buffer will also be called, but not the destructor on m_str since it is a reference.

Note that C++0x changes things just a little bit because it adds rvalue references and move semantics. Essentially by using an rvalue reference parameter (notated with &&) I can 'move' the rvalue into the function (instead of copying it) and the lifetime of the rvalue can be bound to the object it moves into, not the expression. There is a really good blog post from the MSVC team on that walks through this in great detail and I encourage folks to read it.

The pedagogical example for moving rvalue's is temporary strings and I'll show assignment in a constructor. If I have a class MyType that contains a string member variable, it can be initialized with an rvalue in the constructor like so:

class MyType{
   const std::string m_name;
public:
   MyType(const std::string&& name):m_name(name){};
}

This is nice because when I declare an instance of this class with a temporary object:

void foo(){
    MyType instance("hello");
}

what happens is that we avoid copying and destroying the temporary object and "hello" is placed directly inside the owning class instance's member variable. If the object is heavier weight than a 'string' then the extra copy and destructor call can be significant.

-Rick

Rick
+1  A: 

I wrote almost exactly the same class:

http://stackoverflow.com/questions/469696/what-is-your-most-useful-c-c-snippet/473786#473786

Prior to the standard each compiler did it differently. I believe the old Annotated Reference Manual for C++ specified that temporaries should clean up at the end of the scope, so some compilers did that. As late as 2003, I found that behaviour still existed by default on Sun's Forte C++ compiler, so StringBuffer didn't work. But I'd be astonished if any current compiler was still that broken.

Daniel Earwicker
Spooky how similar they are! Thanks for the warning - the first place I'll try it is VC++6, which is not known for its standards compliance. I'll be watching carefully.
Mark Ransom
I would have written the class originally on VC++6 so it shouldn't be a problem.
Daniel Earwicker